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Ta có
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\cdot\frac{5^4.20^4}{25^5.4^5}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\cdot\frac{2^8.5^8}{5^{10}.2^{10}}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\cdot\frac{1}{5^2.2^2}\)
\(=\frac{\left(-2\right)}{6}\cdot\frac{1}{100}=-\frac{1}{3}\cdot\frac{1}{100}=-\frac{1}{300}\)
Vậy : \(E=-\frac{1}{300}\)
Bài làm
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}.\frac{5^4.20^4}{25^5.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}.\frac{5^4.4^4.5^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\frac{5^8.4^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}.\frac{1}{5^2.4}\)
\(\Rightarrow E=\frac{3^8-3^9}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{2}{6}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{1}{300}\)
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.5.3^8}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\frac{2}{6}=-\frac{1}{3}\)
\(E=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.\left[3^8.\left(1-8\right)\right]}{2^{10}.3^8.\left(1+5\right)}=\frac{2^{10}.3^8.\left(-7\right)}{2^{10}.3^8.6}=\frac{-7}{6}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)
\(A=\frac{2^{10}.3^8.\left(-2\right)}{2^{10}.3^8.6}\)
\(A=-\frac{2^{11}.3^8}{2^{10}.3^8.2.3}=-\frac{2^{11}.3^8}{2^{11}.3^9}=-\frac{1}{3}\)
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(E=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.2^2.5}\)
\(E=\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.2^2.5}\)
\(E=\frac{-6}{10}\)
\(E=\frac{-3}{5}\)
vay \(E=\frac{-3}{5}\)
Kết quả cô mik cho là \(\frac{-1}{3}\)cơ !